Question: You have found the following ages (in years) of all 6 snakes at your local zoo: $ 12,\enspace 20,\enspace 16,\enspace 20,\enspace 1,\enspace 17$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{12 + 20 + 16 + 20 + 1 + 17}{{6}} = {14.3\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $12$ years $-2.3$ years $5.29$ years $^2$ $20$ years $5.7$ years $32.49$ years $^2$ $16$ years $1.7$ years $2.89$ years $^2$ $20$ years $5.7$ years $32.49$ years $^2$ $1$ year $-13.3$ years $176.89$ years $^2$ $17$ years $2.7$ years $7.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{5.29} + {32.49} + {2.89} + {32.49} + {176.89} + {7.29}} {{6}} $ $ {\sigma^2} = \dfrac{{257.34}}{{6}} = {42.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{42.89\text{ years}^2}} = {6.5\text{ years}} $ The average snake at the zoo is 14.3 years old. There is a standard deviation of 6.5 years.